# Solve this differential equation explicitly, either by using partial fractions, or with a computer algebra system. Use the initial populations 350 and 450. Can anyone please help me with this problem and explain how you got the answers?

Jun 24, 2018

For the IV: ${P}_{o} = 350$

• $P = 100 \left(\frac{{e}^{\frac{t}{50}} + 5}{{e}^{\frac{t}{50}} - 5} + 5\right)$

For the IV: ${P}_{o} = 450$

• $P = 100 \left(\frac{{e}^{\frac{t}{50}} - 3}{{e}^{\frac{t}{50}} + 3} + 5\right)$

#### Explanation:

$\dot{P} = \frac{P}{10} - {P}^{2} / 10000 - 24$

Complete square:

$= 1 - {\left(P - 500\right)}^{2} / 10000$

Let $Q = P - 500 q \quad \dot{Q} = \dot{P}$

So the DE is:

$\dot{Q} = 1 - {\left(\frac{Q}{100}\right)}^{2}$

Let $R = \frac{Q}{100} q \quad \dot{R} = \frac{\dot{Q}}{100}$

$100 \dot{R} = 1 - {R}^{2}$

This separates:

$\frac{\mathrm{dR}}{1 - {R}^{2}} = \frac{1}{100} \mathrm{dt}$

Factor and use partial fractions:

$= \int \mathrm{dR} \setminus \frac{1}{\left(1 - R\right) \left(1 + R\right)} = \frac{1}{100} \int \setminus \mathrm{dt}$

$= \frac{1}{2} \int \mathrm{dR} \setminus \frac{1}{R + 1} - \frac{1}{R - 1} = \frac{t}{100} + C$

$= \frac{1}{2} \left(\ln \left(R + 1\right) - \ln \left(R - 1\right)\right) = \frac{t}{100} + C$

$= \ln \left(\frac{R + 1}{R - 1}\right) = \frac{t}{50} + C$

$R = \frac{C {e}^{\frac{t}{50}} + 1}{C {e}^{\frac{t}{50}} - 1}$

Reverse the subs:

$Q = 100 \frac{C {e}^{\frac{t}{50}} + 1}{C {e}^{\frac{t}{50}} - 1}$

bb( P= 100 ((Ce^( t/50) + 1)/(Ce^( t/50) - 1) + 5 )

For the IV: ${P}_{o} = 350$

$350 = 100 \left(\frac{C + 1}{C - 1} + 5\right) \implies C = \frac{1}{5}$

$P = 100 \left(\frac{{e}^{\frac{t}{50}} + 5}{{e}^{\frac{t}{50}} - 5} + 5\right)$

For the IV: ${P}_{o} = 450$

$450 = 100 \left(\frac{C + 1}{C - 1} + 5\right) \implies C = - \frac{1}{3}$

$P = 100 \left(\frac{{e}^{\frac{t}{50}} - 3}{{e}^{\frac{t}{50}} + 3} + 5\right)$